Question

establish the identity.
\frac{4\cot u - \tan u}{\cot u+\tan u}+1 = 5\cos^{2}u

write the left side in terms of sine and cosine.
\frac{\square}{\square}+1 (do not simplify.)

Explanation:

Step1: Recall cot and tan definitions

Recall that $\cot u=\frac{\cos u}{\sin u}$ and $\tan u = \frac{\sin u}{\cos u}$.
Substitute these into the left - hand side fraction:
$\frac{4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}}{\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}}+1$

Step2: Get a common denominator for the numerator and denominator of the fraction

The common denominator for the numerator is $\sin u\cos u$, so $4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}=\frac{4\cos^{2}u - \sin^{2}u}{\sin u\cos u}$.
The common denominator for the denominator is $\sin u\cos u$, so $\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}=\frac{\cos^{2}u+\sin^{2}u}{\sin u\cos u}$.
Then the fraction becomes $\frac{\frac{4\cos^{2}u - \sin^{2}u}{\sin u\cos u}}{\frac{\cos^{2}u+\sin^{2}u}{\sin u\cos u}}+1$.
Since dividing by a fraction is the same as multiplying by its reciprocal, $\frac{\frac{4\cos^{2}u - \sin^{2}u}{\sin u\cos u}}{\frac{\cos^{2}u+\sin^{2}u}{\sin u\cos u}}=\frac{4\cos^{2}u - \sin^{2}u}{\cos^{2}u+\sin^{2}u}$.
And we know that $\cos^{2}u+\sin^{2}u = 1$.
So the left - hand side is $4\cos^{2}u-\sin^{2}u + 1$.
Since $\sin^{2}u=1 - \cos^{2}u$, we substitute $\sin^{2}u$:
$4\cos^{2}u-(1 - \cos^{2}u)+1$.

Step3: Simplify the expression

Expand the expression: $4\cos^{2}u-1+\cos^{2}u + 1$.
Combine like terms: $(4\cos^{2}u+\cos^{2}u)+(-1 + 1)=5\cos^{2}u$.

Answer:

The left - hand side in terms of sine and cosine (without simplifying the whole expression after substitution) is $\frac{4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}}{\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}}+1$; and the identity is established as the left - hand side simplifies to $5\cos^{2}u$ which is equal to the right - hand side.